Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

Now I have learnt that! Thanks much Bobbym!

Okay, back to my question.

27^(n+2) =27^n * 27^2 = 3^(3n) * 3^(6)

.

Why didn't you use 3^(3n) * 3^(6) but rather wrote 9^(n+2) * 3^(n+2).

Thanks in advance.

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

All that is correct.

Why didn't you use 3^(3n) * 3^(6) but rather wrote 9^(n+2) * 3^(n+2).

Because it was easier to cancel.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

Candidly speaking, since I started working on indices I haven't seen nor come accros a problem that could produce different bases as your own. As you made 9^(n+2) *3^(n+2) out of 27^(n+2)

Now I think the method that will be applicable to a problem is the one that must be used.

Thank very much Bobbym, God bless you.

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

Hi;

Glad to help and let me know when you need more.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

bobbym wrote:

Okay, I will provide the steps:

Take the log of both sides.

Add log(2) to both sides.

Divide both sides by ( log(2) + log(3) ).

And we are done.

Hi

Initially there was no addition sign[+] but in the course of manipulation you happen to bring an addition sign instead of the multiplication sign.

Please, explain why

thanks

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

Hi;

Line 2?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

Besides, there was two 'x' as in xlog3 + xlog2, And you didn't add those two x to be '2x', or in other words, why didn't you add those x's to be 2x after you divided log2 by log3 + log2, do you understand my question?

I know only one thing - that is that I know nothing

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

bobbym wrote:

Hi;

Line 2?

yes

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

The general rule:

Log(ab) = Log(a) + Log(b)

Under certain conditions of course.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

Please answer my question at #282, if it is unintelligible please tell me

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

Besides, there was two 'x' as in xlog3 + xlog2, And you didn't add those two x to be '2x', or in other words, why didn't you add those x's to be 2x after you divided log2 by log3 + log2, do you understand my question?

Please, what line do you mean?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

See line 4, there is one 'x' behind log2 and another 'x' behind log3, why didn't you add those xs so that it would have been 2x.

Please check these for me:

x( y^1/2 - x^1/2)

= (xy^3/2 - x^3/2)

Is my answer correct?

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

Hi;

I am afraid that

x log(2) + x log(3) does not allow us to add the x's and say 2x.

x( y^1/2 - x^1/2)

= (xy^3/2 - x^3/2)

Is my answer correct?

That is not correct, please try again.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

I think you want me to say; (xy^1/2 - x^1/2)

But I have in mind that the 'x'(the x which is outside of bracket) is raising to the power 1, but the 1 is not clearly written and that I am thinking my answer at 287 is correct, what do you say?

*Last edited by EbenezerSon (2014-08-24 04:17:39)*

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

Hi;

I think you want me to say; (xy^1/2 - x^1/2)

Not exactly, that is incorrect also. You are correct in stating that x is x^1. Now just go through it piece by piece adding the exponents of x.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

I have done that, still I only arrive at the same answer.

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

Hi;

What is x( y^1/2 )?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

To begin with I would say the X instrinsically raised to the power positive one[1], so I will say 1+1/2 = 3/2. Therefore;

XY^3/2

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

But there is no x in there. x^1 = x. What you have is x^0 = 1.

So x( y^1/2 ) = x y^1/2

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

I was talking of the X which is outside the bracket. Look at a similar case at #182 line one, after you had multiplied 2^-x through, the 2 at right side became 2^x+1, this indicates that the 2 already had 1 as its power, and in this problem I perceive that the X has 1 as its power though not visibly expressed or written , what do you say?

Thanks

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

In the parentheses of x( y^1/2 ), there is only a y to the (1/2) power. There is no x in there other than x^0 which equals 1 so we do not include it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

So, if there was X in the bracket having 1/2 as it power how would the answer be? like x(y^1/2 - x^1/2)

Thank you.

*Last edited by EbenezerSon (2014-08-27 04:00:48)*

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

For x(y^1/2 - x^1/2)

you do the first term first:

then the second term:

because you add the exponents which are 1 and 1 / 2 = 3 / 2.

Then you finish off by combining the two terms.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 392

Why did you not say x(y^1/2) = xy^3/2 I think the x has 1 as its power, please I am not getting you with this x(y^1/2) = xy^1/2. Please explicate.

*Last edited by EbenezerSon (2014-08-27 04:49:50)*

I know only one thing - that is that I know nothing

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

Hi;

You can only add the exponents when the bases are the same:

both are x's so add the exponents. both are y's so add the exponents. , it just stays the same. You can do nothing here because the x and y are different.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline